I will mark someone brainliest The snack that smiles back ___

Maureen is making trail mix. She uses 8 ounces of granola and 2 ounces of raisins in her recipe. Fill in each statement. For eve

IrinaK [193]

Answer:

So for every ounce of granola she uses a quarter ounces of raisins so a quarter of 24 is 6

Answer:6

Step-by-step explanation:

4 0

3 years ago

Identify the vertex and the axis of symmetry for y=-(x+3)^2+1

Alex

Vertex
y = -(x + 3)² + 1
y = -(x + 3)(x + 3) + 1
y = -(x² + 3x + 3x + 9) + 1
y = -(x² + 6x + 9) + 1
y = -x² - 6x - 9 + 1
y = -x² - 6x - 8
-x² - 6x - 8 = 0
x = -(-6) +/- √((-6)² - 4(-1)(-8))
 2(-1)
x = 6 +/- √(36 - 32)
 -2
x = 6 +/- √(4)
 -2
x = 6 +/- 2
 -2
x = 6 + 2 x = 6 - 2
 -2 -2
x = 8 x = 4
 -2 -2
x = -4 x = -2
y = -x² - 6x - 8
y = -(-4)² - 6(-4) - 8
y = -(16) + 24 - 8
y = -16 + 24 - 8
y = 8 - 8
y = 0
(x, y) = (-4, 0)
or
y = -x² - 6x - 8
y = -(-2)² - 6(-2) - 8
y = -(4) + 12 - 8
y = -4 + 12 - 8
y = 8 - 8
y = 0
(x, y) = (-2, 0)

Axis of Symmetry
The axis of symmetry is equal to -3.

4 0

4 years ago

In university A, the number of graduate students was approximately 25 comma 000 fewer than the number of undergraduate students

alekssr [168]

Answer:

The number of graduate students attend university A is $G=n-25,000$

Step-by-step explanation:

Consider the provided information.

We need to determine how many graduate students attend university A?

The number of undergraduate students was n.

The number of graduate students was approximately 25,000 fewer than the number of undergraduate students.

Let G represents the number of graduate student.

$G=n-25,000$

Hence, the number of graduate students attend university A is $G=n-25,000$

4 0

3 years ago

Plz with steps plzzzzzz

Stella [2.4K]

Answer: $-\frac{\sqrt{2a}}{8a}$

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

$\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

$\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

$\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\$